I'm suppose to calculate the weight percent per volume (grams pre 100cm3) of ethanoic acid in a commercial sample of vineger from the following: (a) 0.4321g of pure momoprotic acid( formula weight 204.2 amu) required 23.45cm3 of naoh for neutralizalion.

Answer 1

You need to know the molarity of the NaOH.
The 204.2 sounds like you standardized the NaOH by using 0.4321 g of pure potassium hydrogen phthalate.
moles KHP = 0.4321/204.22 = ?
moles NaOH = moles KHP
M NaOH = moles NaOH/0.02435

You can work out the M NaOH.
Now all we need to know is the volume of the vinegar sample you used.

M NaOH x L NaOH = moles NaOH used for the sample.
moles NaOH for the sample = moles acid in the vinegar.
grams acid in the vinegar = moles x molar mass (60).
All of that gives you grams acetic acid/volume used. Convert that to g/100 cc and you have it.
Post your work if you get stuck and PLEASE type ALL of the problem.